Mixing
A little bit confused about a matrix equation
$begingroup$
Find $Ain M_2(mathbb{N} )$ so that $A^2-8A+7I_2=O_2$.
What buggs me is that the eigenvalues of $A$ must satisfy the equation $lambda^2-8lambda+7=0$,so $A$'s eigenvalues are $1$ and $7$. However, $A=I_2$ also satisfies the equation. So what I don't understand is why my approach is wrong.
linear-algebra matrices
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
Find $Ain M_2(mathbb{N} )$ so that $A^2-8A+7I_2=O_2$.
What buggs me is that the eigenvalues of $A$ must satisfy the equation $lambda^2-8lambda+7=0$,so $A$'s eigenvalues are $1$ and $7$. However, $A=I_2$ also satisfies the equation. So what I don't understand is why my approach is wrong.
linear-algebra matrices
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
$endgroup$
$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53
add a comment |
$begingroup$
Find $Ain M_2(mathbb{N} )$ so that $A^2-8A+7I_2=O_2$.
What buggs me is that the eigenvalues of $A$ must satisfy the equation $lambda^2-8lambda+7=0$,so $A$'s eigenvalues are $1$ and $7$. However, $A=I_2$ also satisfies the equation. So what I don't understand is why my approach is wrong.
linear-algebra matrices
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
$endgroup$
Find $Ain M_2(mathbb{N} )$ so that $A^2-8A+7I_2=O_2$.
What buggs me is that the eigenvalues of $A$ must satisfy the equation $lambda^2-8lambda+7=0$,so $A$'s eigenvalues are $1$ and $7$. However, $A=I_2$ also satisfies the equation. So what I don't understand is why my approach is wrong.
linear-algebra matrices
linear-algebra matrices
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
asked Jan 31 at 13:33
JustAnAmateurJustAnAmateur
1096
1096
$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53
add a comment |
$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53
$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53
add a comment |
1 Answer
1
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$begingroup$
Your conclusion is not correct.
If $A^2-8A+7I_2=O_2$, then the minimal polynomial of $A$ divides $lambda^2 - 8lambda + 7$.
So, any matrix with the minimal polynomial $lambda - 1$ or $lambda -7$ will satisfy the above matrix equation.
So, specifically $A=I_2$ will satisfy above equation (as $A = 7I_2$ as well).
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
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1 Answer
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1 Answer
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$begingroup$
Your conclusion is not correct.
If $A^2-8A+7I_2=O_2$, then the minimal polynomial of $A$ divides $lambda^2 - 8lambda + 7$.
So, any matrix with the minimal polynomial $lambda - 1$ or $lambda -7$ will satisfy the above matrix equation.
So, specifically $A=I_2$ will satisfy above equation (as $A = 7I_2$ as well).
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
Your conclusion is not correct.
If $A^2-8A+7I_2=O_2$, then the minimal polynomial of $A$ divides $lambda^2 - 8lambda + 7$.
So, any matrix with the minimal polynomial $lambda - 1$ or $lambda -7$ will satisfy the above matrix equation.
So, specifically $A=I_2$ will satisfy above equation (as $A = 7I_2$ as well).
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
Your conclusion is not correct.
If $A^2-8A+7I_2=O_2$, then the minimal polynomial of $A$ divides $lambda^2 - 8lambda + 7$.
So, any matrix with the minimal polynomial $lambda - 1$ or $lambda -7$ will satisfy the above matrix equation.
So, specifically $A=I_2$ will satisfy above equation (as $A = 7I_2$ as well).
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
$endgroup$
Your conclusion is not correct.
If $A^2-8A+7I_2=O_2$, then the minimal polynomial of $A$ divides $lambda^2 - 8lambda + 7$.
So, any matrix with the minimal polynomial $lambda - 1$ or $lambda -7$ will satisfy the above matrix equation.
So, specifically $A=I_2$ will satisfy above equation (as $A = 7I_2$ as well).
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
edited Jan 31 at 14:02
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
answered Jan 31 at 13:41
trancelocationtrancelocation
13.6k1829
13.6k1829
add a comment |
add a comment |
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$begingroup$
they must satisfy the equation, but necessarily are $1$ and $7$, any combination of these eigenvalues works just fine
$endgroup$
– pointguard0
Jan 31 at 13:39
$begingroup$
The eigenvalues must satisfy the equation. They need not include all the solutions of the equation.
$endgroup$
– Simon
Jan 31 at 13:39
$begingroup$
I think I got it : $A$ need not have both $1$ and $7$ as eigenvalues simultaneously.
$endgroup$
– JustAnAmateur
Jan 31 at 13:53